2016/049) Find the equation of the normal to the curve $x^2= 4y$ that passes through the pont(1,2) (IIT 1984)

equation of the curve $4y = x^2$
so slope of the tangent = $\frac{dy}{dx} = \frac{x}{2}$
so slope of the normal  = $- \frac{2}{x_1}$
the line passes through   $x_1,\frac{x_1^2}{4}$ and $(1,2)$ so slope of line
$\frac{frac{x_1^2}{4} - 2}{x_1-1} = - \frac{2}{x_1}$
$x_1^3 - 8x_1 = - 8x_1 + 8$ or $x_1 = 2$
giving slope of normal = - 1
so equation of normal $= (y-2) = - (x-1)$ or $x+y=3$